LeetCode 532 K-diff Pairs in an Array

题目思路代码

题目

Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array.

A k-diff pair is an integer pair (nums[i], nums[j]), where the following are true:

0 <= i, j < nums.length i != j a <= b b - a == k

Example 1: Input: nums = [3,1,4,1,5], k = 2 Output: 2 Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5). Although we have two 1s in the input, we should only return the number of unique pairs.

Example 2: Input: nums = [1,2,3,4,5], k = 1 Output: 4 Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).

Example 3: Input: nums = [1,3,1,5,4], k = 0 Output: 1 Explanation: There is one 0-diff pair in the array, (1, 1).

Example 4: Input: nums = [1,2,4,4,3,3,0,9,2,3], k = 3 Output: 2

Example 5: Input: nums = [-1,-2,-3], k = 1 Output: 2

Constraints: 1 <= nums.length <= 104 -107 <= nums[i] <= 107 0 <= k <= 107

思路

先将数组排序，对每一个nums[i]去右侧寻找是否有数可以构成k-pair，如果出现差值已经大于k的情况直接break（因为后面的数更大）。while (i >= 1 && i < nums.length && nums[i] == nums[i - 1]) i++;是为了处理重复元素的时候只有第一次计算，后面跳过，防止多算。

代码

class Solution { public int findPairs(int[] nums, int k) { int res = 0; Arrays.sort(nums); for (int i = 0; i < nums.length; i++) { while (i >= 1 && i < nums.length && nums[i] == nums[i - 1]) i++; for (int j = i + 1; j < nums.length; j++) { if (nums[j] - nums[i] == k) { res++; break; } if (nums[j] - nums[i] > k) break; } } return res; } }